Question

If the magnitude of vectors$\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{C}$ are 12, 5 and 13, respectively and $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$, then find the angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ .

Answer 1

Hint: In our question, we can use the properties of vectors and the properties of a triangle to find the desired answer. We can check whether the given triangle has any specific feature and if yes, we can use trigonometric formulas to find the answer.

Step by step answer:We can approach this question in two methods. One is by simple geometry and the other is using simple vector properties.
When we consider our first method, we can say that we are given with a right angle triplet in our question, which is 12, 5 and 13. Upon summing the squares of 12 and 5, we obtain the square of thirteen, that is $144+25=169$.
Hence, we can say that $\overrightarrow{C}$is the hypotenuse and also, $\overrightarrow{A}$ and $\overrightarrow{B}$are the other two sides of the right triangle, which forms an angle of ${{90}^{\circ }}$between them.
Using the vector method, we are given that
$\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$
Upon squaring it , or taking its dot product with itself, we obtain
$ \overrightarrow{C}.\overrightarrow{C}=\overrightarrow{(A}+\overrightarrow{B}).\overrightarrow{(A}+\overrightarrow{B}) \\
\Rightarrow {{C}^{2}}={{A}^{2}}+{{B}^{2}}+2AB\cos \theta \\
$
Upon substituting the given values from the question, we get
$
  {{13}^{2}}={{12}^{2}}+{{5}^{2}}+2(12)(5)\cos \theta \\
  \Rightarrow \cos \theta =0 \\
  \therefore \theta ={{90}^{\circ }} \\
$
Hence, $\overrightarrow{A}$ and $\overrightarrow{B}$ forms an angle of ${{90}^{\circ }}$between them.

Note: Dot products are also known as scalar products because this operation takes up two vectors and returns a scalar value after operation. It can be defined as the product of magnitude of two vectors and the cosine of the angle between the two angles. Geometrically it is the Euclidean product of the vectors and its cosine.